2 Unknowns Curve Sketching (1 Viewer)

[Serb4Lyf]

Sketch the following functions showing the maximum and minimum points on the graph:<o></o>

How do you do it with 2 unknowns?

 

[Slidey]

Huh? You can do it with a function in 2 unknowns - you usually graph trivial and extreme cases (indicating the range in between) or you use field vectors and stuff. That's assuming one of the unknowns is arbitrary whilst the other is variable. If they are both variable you use multivariate calculus.

But those are all your typical functions of one unknown.

1) Find x and y intercepts (when x=0 and y=0, so use quadratic formula to find roots)
2) Take the derivative to find turning points and rate of change (but they are all parabola's so they just look like a U )

E.g. the first one, a)
y=(x-2)^2

The parabola touches the x axis at x=2. Because the parabola is completely positive (any number suqared is positive), the entire graph is positive. That is y >= 0 for all x.

Basically, y approaches infinity for both positive and negative values of x, and has a minimum value at x=2 (we know this because it only has one root, thus the root is the maximum or minimum point, depending on the sign of the graph).

 

[midifile]

Sketch the following functions showing the maximum and minimum points on the graph:<o>></o>>

How do you do it with 2 unknowns?

Seriously? All functions that arent staight horizontal or vertical lines have two 'unknowns'. If you are confused because the question says they are functions but there is no f(x) replace the y with a f(x) they mean the same thing.

a) y = x^2 - 4x +4
= (x-2)^2
Therefore this is a normal parabol moved two units to the right
You can also use differentiation if you have learnt it to workout the maximum or minimum. If not, you can work out the axis of symmetry by doing -b/2a = 4/2 = 2. This is the x value of the maximum/minimum (minimum in this case).
Then sub this into the equation to find the y value: y=0
Therefore the minimum is (2,0)

b) y= x^2 - 7x +12
= (x-4)(x-3)
This is parabola that goes through 3 and 4 on the x axis and 12 on the y axis
Likewise, the axis of symetry can be worked out by doing -b/2a
therefore x= 7/2
y = (7/2)^2 -7(7/2) + 12
= -1/4
Minimum = (7/2, -1/4)

c) For this one make the x axis the t axis
y=3t^2 - 12t + 12
=3(t^2 - 4t + 4)
=3(t-2)^2
So this is a parabola moved two units the the right and 3 times as steep.
axis of symmetry = 12/6
= 2
y value = 3(2)^2 - 12(2) +12
= 0
Therefore minimum=(2,0)